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Example 11 - Chapter 3 Class 11 Trigonometric Functions (Term 2)
Last updated at Nov. 17, 2020 by Teachoo
Transcript
Example 11 Find the value of sin 15°. sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° = 1/√2 × √3/2 −1/√2 × 1/2 = 1/√2 ((√3 − 1)/2) = (√𝟑 − 𝟏)/(𝟐√𝟐) sin (x – y) = sin x cos y – cos x sin y Putting x = 45° y = 30°
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Chapter 3 Class 11 Trigonometric Functions (Term 2)
Concept wise
- Radian measure - Conversion
- Arc length
- Finding Value of trignometric functions, given other functions
- Finding Value of trignometric functions, given angle
-
(x + y) formula
- 2x 3x formula - Proving
- 2x 3x formula - Finding value
- cos x + cos y formula
- 2 sin x sin y formula
- Finding Principal solutions
- Finding General Solutions
- Sine and Cosine Formula
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.